T.3 · Комплексный анализ и вычеты
GRE Subject Test · GRE Mathematics Тема 7 2:40 Английское озвучивание · Английский + китайские субтитры (встроенные)
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This unit tests complex differentiability, uses contour integrals and residues, and evaluates contour integrals with simple and higher-order poles.
本单元检验复可微性，使用围道积分和留数，并计算含单极点和高阶极点的围道积分。
Write f as u plus i v.
把 f 写成 u + iv。
Complex differentiability imposes u x equals v y and u y equals minus v x; with continuous partials nearby, these establish analyticity.
复可微要求 uₓ = v_y 且 u_y = −vₓ；若附近偏导数连续，这就保证了解析性。
z squared passes everywhere.
z² 处处满足。
The conjugate fails everywhere, although its real partials exist.
共轭函数处处不满足，尽管它的实偏导数存在。
A removable singularity can be filled when the function is bounded near it.
可去奇点在函数于其附近有界时可以补上。
A pole has finitely many negative powers; an essential singularity has infinitely many.
极点只有有限个负幂项；本性奇点有无穷多个。
The residue is the coefficient of z minus a to the minus 1, not necessarily the leading term.
留数是 (z − a)^{−1} 的系数，不一定是最高的负幂项。
For a pole of order m, write f as g over z minus a to the m; the residue is the m minus first derivative of g at a, over m minus 1 factorial.
对 m 阶极点，把 f 写成 g/(z − a)^m；留数是 g 在 a 处的 m − 1 阶导数除以 (m − 1)!。
So e to the z over z minus a squared has residue e to the a.
所以 e^z/(z − a)² 的留数是 e^a。
The residue theorem then gives 2 pi i times the enclosed residues, for a positively oriented contour.
留数定理给出：沿正向围道的积分等于 2πi 乘以内部留数之和。
For the conjugate, u x is 1 but v y is minus 1, so it is not complex differentiable.
对共轭函数，uₓ = 1 而 v_y = −1，所以不是复可微的。
For 1 over z minus 2, the circle of radius 1 about 2 encloses a simple pole of residue 1, so the integral is 2 pi i.
对 1/(z − 2)，以 2 为圆心、半径为 1 的圆包含一个留数为 1 的单极点，所以积分是 2πi。
Continuity or real differentiability alone does not imply complex analyticity.
仅有连续性或实可微性，并不能推出复解析性。
Expand e to the z as 1 plus z plus z squared over 2.
把 e^z 展开为 1 + z + z²/2 + …。
Dividing by z squared, the coefficient of z to the minus 1 is 1.
除以 z² 后，z^{−1} 的系数是 1。
So the residue is 1 and the integral is 2 pi i: a double pole does not force a zero residue.
所以留数是 1，积分是 2πi：二阶极点不一定留数为零。
Only zero is inside the unit circle, with residue the limit of 1 over z minus 2: minus a half.
单位圆内只有零点，其留数是 1/(z − 2) 的极限：−1/2。
Counterclockwise the integral is minus pi i; clockwise, pi i.
逆时针积分为 −πi；顺时针为 πi。
The pole at 2 is outside and contributes nothing.
2 处的极点在外面，没有贡献。
But the circle of radius 2 passes through a pole, so the theorem cannot be used there.
但半径为 2 的圆经过一个极点，所以那里不能用留数定理。
Check yourself.
自我检测。
The conjugate is nowhere analytic.
共轭函数处处不解析。
e to the z over z squared has residue 1 at zero.
e^z/z² 在零点的留数是 1。
A constant over a pure double pole has residue zero.
常数除以纯二阶极点，留数为零。
A clockwise contour flips the sign.
顺时针围道改变符号。
And a pole on the contour stops the theorem.
围道上有极点时定理不适用。